The student council plans to have a BBQ lunch for the whole school to celebrate its victory in the volleyball finals. They need a total of 427 hotdogs and hamburgers. Hot dogs costs $0.25 each and hamburgers cost $0.80 each they have a budget of $203 for the event. How many of each should the school council buy?
number of hotdogs --- x
number of hamburgers -- (427-x)
.25x + .8(427-x) = 203
times 100
25x + 80(427-x) = 20300
-55x + 34160 = 20300
-55x = -13860
x = 252
need 252 hotdogs and 175 hamburgers
Let's assume the number of hotdogs the student council should buy is "x", and the number of hamburgers they should buy is "y".
According to the given information, the total number of hotdogs and hamburgers needed is 427, so we can write the equation:
x + y = 427 (Equation 1)
The cost of each hotdog is $0.25 and each hamburger is $0.80, and the budget available is $203. Thus, the cost equation can be written as:
0.25x + 0.80y = 203 (Equation 2)
Now, we can solve these two equations to find the values of "x" and "y".
Using substitution method, we can solve Equation 1 for x:
x = 427 - y
Substituting this value of x in Equation 2:
0.25(427 - y) + 0.80y = 203
107 - 0.25y + 0.80y = 203
0.55y = 96
y = 96 / 0.55
y ≈ 174.55
Since we can't have a fractional part of hamburgers, we need to round it down to the nearest whole number. Therefore, y ≈ 174.
Substituting this value of y back into Equation 1:
x + 174 = 427
x = 427 - 174
x = 253
Therefore, the student council should buy approximately 253 hotdogs and 174 hamburgers to stay within their budget and meet the required quantity.
To determine how many hotdogs and hamburgers the student council should buy, we'll use a system of equations.
Let's represent the number of hotdogs as "H" and the number of hamburgers as "B."
Given information:
- The total number of hotdogs and hamburgers needed is 427.
- The cost of each hotdog is $0.25, and the cost of each hamburger is $0.80.
- The budget for the event is $203.
Now, we can set up the following equations:
1) H + B = 427 (equation representing the total number of hotdogs and hamburgers needed)
2) 0.25H + 0.80B = 203 (equation representing the budget limit)
To solve this system of equations, we can use the substitution or elimination method.
Let's solve it using the substitution method:
From equation 1), we can express H in terms of B by subtracting B from both sides:
H = 427 - B
Substitute this expression for H into equation 2):
0.25(427 - B) + 0.80B = 203
Simplify and solve for B:
106.75 - 0.25B + 0.80B = 203
0.55B = 96.25
B = 96.25 / 0.55
B ≈ 175
Now, substitute the value of B back into equation 1):
H + 175 = 427
H = 427 - 175
H = 252
Therefore, the student council should buy 252 hotdogs and 175 hamburgers for the BBQ lunch.